Linear approximation
Problem 3.188 · easy
Use the linear approximation of \( \displaystyle f(x) = \sqrt{x} \) at \( \displaystyle x = 16 \) to estimate \( \displaystyle f(\frac{84}{5}) \).
- The linearization at a is L(x) = f(a) + f'(a)(x − a).Reviewed
- \[ \left. \sqrt{x} \right|_{\substack{ x=16 }} = 4 \]f(a).✓ Proved
- \[ \left. \frac{d}{d x} \sqrt{x} \right|_{\substack{ x=16 }} = \frac{1}{8} \]f'(a).✓ Proved
- \[ \frac{x}{8} + 2 \]The linearization.✓ Proved
- \[ \frac{41}{10} \]Evaluate at x = \frac{84}{5}.✓ Proved
Answer \( L(\frac{84}{5}) = \frac{41}{10} \approx 4.1 \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the estimate is within Taylor's error bound of the true value |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the linear approximation formula, computes the derivative and function value at the point of tangency, and evaluates the linearization at the target point. The arithmetic is correct.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies the linear approximation formula, computes the derivative and function value at the point of tangency, and evaluates the linearization at the target point. The arithmetic is correct.gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies the linear approximation formula, computes the derivative and function value accurately, and evaluates the linearization at the specified point.gpt-oss:20b: pass 2026-09-26
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/linear_approximation, checked 2026-09-26 with SymPy 1.14.0.